# Why don't homophones hide multiple-letter patterns?

I am reading about mono-alphabetic ciphers, which are prone to letter frequency analysis. To counter this, we can provide multiple substitutions, known has homophones for a single letter e.g. e could be assigned 17, 74, 35, 21. Each homophone could be assigned in rotation or randomly.

This is where I am confused:

Even with homophones, each element of plaintext affects only one element of ciphertext, and multiple-letter patterns (e.g. diagram frequencies) still survive in the ciphertext, making cryptanalysis relatively straight forward.

How do multiple-letter patterns survive? For example if for h we have homophones 17, 74, 35, 21 and for t we have 11, 69, 27, 24, th could be 11 17 or 11 74 or 11 35 or 69 21 etc.

• Surely the qualtity of the frequency distributions of the encrypted pairs (corresponding to your digrams) will depend on the details of your homophone scheme. BTW, I had earlier proposed elsewhere a digram transformation (that doubles text length!) which employs three Playfair matrices placed in the form of an L. One first encrypts (p1,p2) into (c1,c2) with the matrix at the root of L. Then c1 is turned into (h11,h12) where h11 is any character in the upper matrix in the same column as c1 and h12 is any character in the right matrix in the same row as c1. Similarly c2 is turned into (h21,h22). Sep 13 '16 at 15:07

For starters, you don't have 4 or 5 codes for every letter. The basic idea is, that each symbol in your alphabet for the ciphertext has the same frequency. Most letters will still have only 1 or 2 symbols, and just the more common ones will have 3 or more.

So the frequencies of single symbols is roughly uniform. Let's take your example of a th, which is the most common bigram in english language (the frequencies are just examples, other sources might have different ones)

• $t$ has frequency $0.0894$
• $h$ has a frequency $0.0496$
• $th$ has a frequency of $0.0271$

Now let's imagine we have a total of 100 symbols in the ciphertext, where we have 9 symbols for the letter $t$ and 5 for the letter $h$, so that each symbol has roughly the frequency $1/100$.

Now there are $9\cdot5 = 45$ possible combinations for $th$. With 100 symbols in the alphabet, the number of possible bigrams is $10000$ and the frequency should be 0.0001 if it is uniform. We know that $th$ has a frequency of $0.0271$, spreding that uniformly over 45 combinations and we get a frequency of $0.0271 / 45 \approx 0.0006$.

So, even if we have the monogram probabilities almost at $1/100$ for every symbol, then those $45$ bigrams for $th$ have a frequency that is roughly 6-times as high as the average bigram should be. In comparison: The monogram $t$ has just a frequency, which is not even 2.5 times higher than $1/26 \approx 0.0385$.